4v2+16v=65

Solution for 4v2+16v=65 equation:

4v^2+16v=65

We move all terms to the left:

4v^2+16v-(65)=0

a = 4; b = 16; c = -65;
Δ = b2-4ac
Δ = 162-4·4·(-65)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-36}{2*4}=\frac{-52}{8}
=-6+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+36}{2*4}=\frac{20}{8}
=2+1/2
$